similar to: covariates in lm

Is it possible to use R as a data-mining tool? Here's the problem I've got. I have a couple of data sets consisting of results from a cDNA microarray experiment - the details about the biology don't really matter here, the same theory applies for any other data-mining task (that's why I thought it'd be more appropriate to post this on r-user). Each of these datasets consists

Hmm. Chuck: I don't see how this example represents incomplete/incommensurate recycling. Doesn't TRUE replicate from length-1 to length-3 in this case (mat[c(TRUE,FALSE),2] would be an example of incomplete recycling)? William: clever, but maybe too clever unless you really need the speed? (The clever way is 8 times faster in the following case ...) x <- rep(1,1e6)

First, thanks to all who helped me with my question about rescaling axes on the fly. Using unlist() and range() to set the axis ranges in advance worked well. I've since plotted about 300 datasets with relative ease. Now I'm trying to fit a lossy oscillator resonance to (the square root of) a lorentzian (testframe$y is oscillator amplitude, testframe$x is drive frequency): lorentz

Hi, 2 questions: Question 1: example of what I currently do: for(i in 1:6){sink("temp.txt",append=TRUE) dput(i+0) sink()} x=scan(file="temp.txt") print(prod(x)) file.remove("C:/R-2.5.0/temp.txt") But how to convert the output of the loop to a vector that I can manipulate (by prod or sum etc), without having to write and append to a file? Question 2: >

Hi, I'm fitting a standard nonlinear model to the luminances measured from the red, green and blue guns of a TV display, using nls. The call is: dd.nls <- nls(Lum ~ Blev + beta[Gun] * GL^gamm, data = dd, start = st) where st was initally estimated using optim() st $Blev [1] -0.06551802 $beta [1] 1.509686e-05 4.555250e-05 7.322720e-06 $gamm [1] 2.511870 This works fine but I

Dear list-subscriber, in the process of writing a general code snippet to extract coefficients in an expression (in the example below: 0.5 and -0.7), I stumbled over the following peculiar (at least peculiar to me:-) ) sorting behaviour of the function all.names(): > expr1 <- expression(x3 = 0.5 * x1 - 0.7 * x2) > all.names(expr1) [1] "-" "*" "x1"

Hello! I work with: R : Copyright 2006, The R Foundation for Statistical Computing Version 2.3.1 (2006-06-01) On Windows XP Professional (Version 2002) SP2 I think there is a bug in the conditional execution if (expr1) {expr2} else {expr3} If I try: "if (expr1) expr2 else expr3" it works well but when I put the expression expr2 and expr3 between {} I receive an error message

This may have been asked before, but is there an elegant way to check whether an variable/argument passed to a function is a "parse tree" for an (unevaluated) expression or not, *without* evaluating it if not? Currently, I do various rather ad hoc eval()+substitute() tricks for this that most likely only work under certain circumstances. Ideally, I'm looking for a isParseTree()

Hi, Is there any equivalent for ifelse (except if (cond) expr1 else expr2) which takes an atomic element as argument but returns vector since ifelse returns an object of the same length as its argument? x = c(1,2,3) y = c(4,5,6,7) z = 3 ifelse(z <= 3,x,y) would return x and not 1 thanks

Hello, I have a function for reading a data-frame from a file, which contains E = read.table(file = filename, header = T, colClasses = c(rep("integer",6),"numeric","integer",rep("numeric",8)), ...) Now a small variation arose, where colClasses =

Dear list # I have a function ff <- function(a,b=2,c=4){a+b+c} # which I programmatically want to modify to a more specialized function in which a is replaced by 1 ff1 <- function(b=2,c=4){1+b+c} # I do as follows: vals <- list(a=1) (expr1 <- as.expression(body(ff))) expression({ a + b + c }) (expr2 <- do.call("substitute", list(expr1[[1]], vals))) { 1 +

I am more than a little puzzled by your question. In the construct {expr1; expr2; expr3} all of the expressions expr1, expr2, and expr3 are evaluated, in that order. That's what curly braces are FOR. When you want some expressions evaluated in a specific order, that's why and when you use curly braces. If that's not what you want, don't use them. Complaining about it is like

hi , this can be done easily if (cond) expr ex:  > for (i in 1: 4)+ {+ if(i==2) print("a")+ if(i==2) print("b")+ } output : [1] "a"[1] "b" but i want this  if (cond) expr1 expr 2 i tried this :  > for (i in 1: 4)+ {+ if(i==2) (print("b") && print("a"))+ } output : [1] "b"Error in print("b") &&

Today I ran into another aspect of the poison problem... Basically, SimplifyCFG wants to take expr1 && expr2 and flatten it into x = expr1 y = expr2 x&y This isn't safe when expr2 might execute UB. The consequence is that no LLVM shift instruction is safe to speculatively execute, nor is any nsw/nuw/exact variant, unless the operands can be proven to be in

R's { expr1; expr2; expr3} acts much like C's ( expr1, expr2, expr3) E.g., $ cat a.c #include <stdio.h> int main(int argc, char* argv[]) { double y = 10 ; double x = (printf("Starting... "), y = y + 100, y * 20); printf("Done: x=%g, y=%g\n", x, y); return 0; } $ gcc -Wall a.c $ ./a.out Starting... Done: x=2200, y=110 I don't like that

Hi >i have aproblem withe execution of my function >first, i wrote my function in the script of R >nom_fonction <- function(arg1[=expr1], arg2[=expr2], ...){ bloc d'instructions } > when i want to have the result i mean the laste instruction in the bloc of > instruction , i try to >wrote the name of function >source(aj.fun) Error in readLines(file, warn =

Hi all, I am not used to using the computer to do calculus and have up to now done my differentiation "by hand" , calling on skills I learned many years ago and some standard cheat sheets. My interest at present is in getting the second derivative of a gaussian, which I did by hand and results in a somewhat messy result involving terms in sigma^5 .. I have done some spot checks

Dear List, I am using constrOptim to solve the following fr1 <- function(x) { b0 <- x[1] b1 <- x[2] ((1/(1+exp(-b0+b1))+(1/(1+exp(-b0)))+(1/(1+exp(-b0-b1)))))/3 } As you can see, my objective function is ((1/(1+exp(-b0+b1))+(1/(1+exp(-b0)))+(1/(1+exp(-b0-b1)))))/3 and I would like to solve for both b0 and b1. If I were to use optim then I would derive the gradient of the

Sorry if this has been covered here somewhere in the past, but ... Does anyone know why logical vectors are *silently* recycled, even when they are incommensurate lengths, when doing logical indexing? This is as documented: For ?[?-indexing only: ?i?, ?j?, ?...? can be logical vectors, indicating elements/slices to select. Such vectors are recycled if necessary to match

Hey, everybody--- Ignorance CAN be bliss, at least for a while, but, .... Just so you know... A week or two ago, some upgrades to the expression parser (you know, the expressions you put in $[ ... ] in your extensions.conf file) that I submitted, have been merged into the CVS HEAD of the source. Hopefully, for around 99.9% of you, it won't make any difference to you. The Makefile has also